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Consider $\mathbb{R}^{n}$ with natural topology. Let $K\in \mathbb{R}^{n} $ compact set and $K \subset \bigcup _{n\in \mathbb{N}}K(r_{n},1)$, where $K(r_{n},1)$ is open ball (in Euclidean metric). Show that exists $r\in \mathbb{R}^{n}$ and $\delta>0$ such as $K\subseteq K(r,\delta)$

My solution is:

So if $K$ is compact, then:

1) $K$ is Hausdorff space,

2) from every open cover there is a finite subcover.

Now, $\bigcup _{n\in \mathbb{N}}K(r_{n},1)$ is finite cover of $K$. As we know, $K$ is compact, so there exists finite subcover of $\bigcup _{n\in \mathbb{N}}K(r_{n},1)$, $r\in \mathbb{R}^{n}$ and $\delta>0: K(r,\delta)$

We concude that:

$K\subset \bigcup _{n\in \mathbb{N}}K(r_{n},1) \subset K(r,\delta) $

Am I correct?

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  • $\begingroup$ $K\color{red}{\in}\Bbb R^n$ is incorrect. $\endgroup$ – Chris Custer Jan 23 at 1:17
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You did not say what $r$ and $\delta$ are. Take $r=0$ and $\delta=1+\max \{\|r_i\|: 1\leq i \leq n\}$. Note that $x \in K$ implies $\|x-r_i\| <1$ for some $i$. This gives $\|x-0\| \leq \|r_i\|+\|x-r_i\|<1+\|r_i\| <\delta$.

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  • $\begingroup$ Can I take $r=0$? Such ball does not exist? $\endgroup$ – Michal Jan 22 at 23:52
  • $\begingroup$ @Michal By $r=0$ I mean $r =(0,0,...,0)$. This vector belongs to $\mathbb R^{n}$ so $K(r,\delta)$ exists for any $\delta >0$. $\endgroup$ – Kavi Rama Murthy Jan 22 at 23:54
  • $\begingroup$ Forgive me, for a second I thought $r$ was radius... Thanks for clarification. $\endgroup$ – Michal Jan 22 at 23:57

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